Econ 805 Problem Set 3 Solution
1.
(Gibbons 1.2).
In the following game, what strategies survive IESDS (iterated
elimination of strictly dominated strategies)? What are the pure strategy Nash equilibria?
L
C
R
T
2
,
0
1
,
1
4
,
2
M
3
,
4
1
,
2
2
,
3
B
1
,
3
0
,
2
3
,
0
Answer: for player 1,
B
is strictly dominated by
T
. So we can delete the row of
B
.
Then for player 2,
C
is strictly dominated by
R
. No further strategies can be deleted.
Therefore, the strategies that survive IESDS are:
(
T, M
)
and
(
L, R
)
.
The pure strategy NE of this game are
(
M, L
)
and
(
T, R
)
(see the underscore in the
bimatrix).
2. Mascolell 8.B.5.
Answer: (i) First, any quantity
q > q
m
=
a
−
c
2
b
(the monopoly quantity) is strictly
dominated by
q
m
.
Then the remaining strategy set for each
fi
rm is
[0
, q
m
]
.
Now you
can calculate the best response to
q
m
. You will get
R
(
q
m
) =
a
−
c
2
b
−
q
m
2
.. One can show
that any
q < b
(
q
m
)
is strictly dominated by
R
(
q
m
)
.
Now the relevant strategy set is
[
R
(
q
m
)
, q
m
]
. Now you can calculate the best response to
R
(
q
m
)
. You can easily get that
R
2
(
q
m
) =
a
−
c
2
b
−
R
(
q
m
)
2
.
Again you can prove that any quantity
q > R
2
(
q
m
)
is strictly
dominated by
R
2
(
q
m
)
. Now the relevant strategy set shrinks to
[
R
(
q
m
)
, R
2
(
q
m
)]
. If you
continue this process, after
2
n
rounds of IESDS, the remaining strategy set is
[
R
2
n
−
1
(
q
m
)
,
R
2
n
(
q
m
)]
. Now you only need to show that
R
n
(
q
m
)
converges as
n
goes to in
fi
nity. You
can show that
R
n
(
q
m
)
=
2
n
−
2
n
−
1
+ 2
n
−
2
+
...
+ (
−
1)
n
2
n
−
n
2
n
q
m
=
[1
−
1
2
+
1
4
−
...
+ (
−
1)
n
1
2
n
]
q
m
=
(
{
1
−
1
4
[1 +
1
4
+
...
+
1
4
n
2
−
1
]
}
q
m
if
n
is even
1
2
[1 +
1
4
+
..
+
1
4
n
−
1
2
]
q
m
if
n
is odd
If
n
is even,
R
n
(
q
m
) =
{
1
−
1
4
1
−
(
1
4
)
n
2
−
1
1
−
1
4
}
q
m
→
2
3
q
m
If
n
is odd,
R
n
(
q
m
) =
1
2
1
−
(
1
4
)
n
−
1
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 Spring '08
 zeca
 Gibbons

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